If a stereo has a dial that changes the volume of sounds the stereo makes what is the dial doing

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$

this expression holds for the points located at

-r₁ <x₁ <r₁

3) Point between the two spheres

E_ {total} = $k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}$

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

E_ {total} = $+ k \frac{Q}{(d-x_1)^2}$ + k Q / (d-x1) 2

point range

-r₂ <x₂ <r₂

5) Right side point

E_ {total} = $k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}$

E_ {total} = $- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )$ - k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0

2 years ago

A 4.00-kg model rocket is launched, expelling 64.0 g of burned fuel from its exhaust at a speed of 675 m/s. What is the velocity

olga_2 [115]

Answer:

The velocity of the rocket is 7.8 m/s

Explanation:

3 0

2 years ago

Imagine Two Artificial Satellites Orbiting Earth At The Same Distance. One Satellite Has A Greater Mass Than The Other One? Whic

Bad White [126]

After reading this whole question, I feel like I've already
earned 5 points !

-- Two satellites at the same distance, different masses:

The forces of gravity between two objects are directly
proportional to the product of the objects' masses. In
other words, the gravitational forces between the Earth
and an object on its surface are proportional to the mass of
the object. In other words, people with more mass weigh more
on the Earth, and the Earth weighs more on them.

If the satellites are both at the same distance from Earth,
then the Earth pulls on the one with more mass with greater
force, and also the one with more mass pulls on the Earth
with greater force.

-- Two satellites with the same mass, at different distances:

The forces of gravity between two objects are inversely
proportional to the square of the distance between them.
In other words, the gravitational forces between the Earth
and an object are inversely proportional to the square of
the distance between the object and the center of the Earth.

If the satellites both have the same mass, then the Earth
pulls on the nearer one with greater force, and also the
nearer one pulls on the Earth with greater force.

-- Resistor in a circuit when the voltage changes:

The resistance depends on how the resistor was manufactured.
Its resistance is marked on it, and doesn't change. It remains
the same whether the voltage changes, the current changes,
the time of day changes, the cost of oil changes, etc.

If you increase the voltage in the circuit where that resistor is
installed, the current through the resistor increases. If the current
remains constant, then you can be sure that somebody snuck over
to your circuit when you weren't looking, and they either installed
another resistor in series with the original one to make the total
resistance bigger, or else they snipped the original one out of the
circuit and quickly connected one with more resistance in its place.

6 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

What is the force exerted on a charge of 2. 5 µC moving perpendicular through a magnetic field of 3. 0 × 102 T with a velocity o

stira [4]

The force acting on a moving charge is known as the magnetic force. The force acting on the charge will be 3.75 N.

<h3>What is the force exerted on the charge?</h3>

Magnetic fields only exert a force on a moving electric charge. A moving charge generates a magnetic field. With an increase in charge and magnetic field strength, this force rises.

when charges have higher velocities, the force is stronger. However, the magnetic force is always perpendicular to the velocity.

Mathematically the force exerted on the charge will be

F=qvBsinα

F= force acting on the charge

v = velocity of charge

q = charge

F=qvBsinα

F=2.5×10⁻⁶×5.0×10³×3.0×10²

F=37.5 N

Hence The force acting on the charge will be 3.75 N.

To learn more about the force acting on charge refer to ;

brainly.com/question/451411

F = q V B sinα

Where F is the force applied to a moving charge.

V = charge velocity

q stands for charge.

α = angle between V and B directions

As a result, the moving charge is subjected to a force of 3.75 Newton.

3 0

1 year ago